Hmm, I didn't care much about the collision of marathon contest and algorithm contest, but it seems it affected the international participants a lot. We'll avoid the collision in the future.

1. Observe that if the first character of S is A then it becomes B, otherwise everything flip, and we delete the first element, and add a 'A' at the end.

2. Convince yourself that the sequence becomes more or less static very fast.

Uploaded.

The point is that the contribution of each and every two connected component differs by whether it is a bipartite graph. Leave the single point alone, in a single connected component, you can color the graph with black & white (two side of each edge have different colors). If possible, pair (black, white) will attach to a (white, black) with an edge, and it cannot connect with point like pair (white, white) and (black, black).

Likewise, you can use similar way to calculate the contribution of two distinct connected components.

2X points at Atcoder = X points at Topcoder Div1

I think if you get (or somehow already know) the reunity number fact it's much easier to proceed, but without that D is definitely easier than E.

Yes, I agree that F's partial score is much harder than 500 points, but it's somewhat difficult to give partial points. For example, if we give 1200 points for it, the difference between the full score and the partial score is also much bigger than 500 points.

If you can represent n as sum of k powers of 10 then you can do it also using k+9 powers of 10 (if that's not greater than n). Moreover 9n+9k is divisible by 9, so its sum of digits is also divisible by 9.